Fundamental Theorem of Calculus Worksheet

Use the fundamental theorem of calculus, to evaluate

\int_{1}^{2}

(

x

⁴ - 3

x

)

dx

	a.		$\frac{9}{2}$
	b.		$\frac{17}{10}$
	c.		$\frac{33}{5}$
	d.		$\frac{107}{10}$
	e.		$\frac{31}{5}$

Solution:

∫12(x⁴ - 3x) dx

= ∫12x⁴ dx - 3 ∫12x dx
[Use Difference Rule of integration.]

= [x55]12 - 3 [x22]12

= [325-15] - 3 [4 / 2 - 1 / 2]
[Apply the limits of integration.]

= 31 / 5 - 9 / 2

= 62-4510 = 17 / 10

So, ∫12(x⁴ - 3x) dx = 17 / 10

Correct answer : (2)

Use the fundamental theorem of calculus, to evaluate

\int_{0}^{3}

\frac{1}{4 x + 3}

dx

	a.		$\frac{1}{4}$ ln 18
	b.		$\frac{1}{4}$ ln 5
	c.		ln 5
	d.		$\frac{1}{4}$ ln 3
	e.		$\frac{1}{3}$ ln 5

Solution:

∫0314x+3 dx

= 1 / 4 [ln | 4x + 3 | ]03

= 1 / 4 [ln 15 - ln 3]
[Apply the limits of integration.]

= 1 / 4 ln(15 / 3)

= 1 / 4 ln 5

So, ∫0314x+3 dx = 1 / 4 ln 5

Correct answer : (2)

Use the fundamental theorem of calculus to evaluate,

\int_{0}^{1}

(

e

^{3 $x$} +

e

^{-3 $x$})

dx

	a.		$\frac{e^{3}}{3}$
	b.		$\frac{e^{3} - 1}{e^{3}}$
	c.		$\frac{e^{3} + 1}{3 e^{3}}$
	d.		$\frac{e^{6} + 1}{3 e^{3}}$
	e.		$\frac{e^{6} - 1}{3 e^{3}}$

Solution:

∫01(e^3x + e ^-3x) dx

= ∫01e^3x dx + ∫01e ^-3x dx
[Use Sum Rule of integration.]

= [e3x3]01 + [e-3x-3]01

= e33 - 1 / 3 + e-3-3 + 1 / 3
[Apply the limits of integration.]

= e33-e-33

= e3-e-33 = e6-13e3

So, ∫01(e^3x + e ^-3x) dx = e6-13e3

Correct answer : (5)

Use fundamental theorem of calculus, to evaluate

\int_{0}^{π/3}

(2 sin

x

+ 3 cos

x

)

dx

	a.		$\frac{2 - 3 \sqrt{3}}{2}$
	b.		$\frac{1 + \sqrt{3}}{2}$
	c.		$\frac{\sqrt{3}}{2}$
	d.		$\frac{2 + 3 \sqrt{3}}{2}$
	e.		$\frac{1 - \sqrt{3}}{2}$

Solution:

∫0π/3(2 sin x + 3 cos x) dx

= 2 ∫0π/3sin x dx + 3 ∫0π/3cos x dx
[Use Sum Rule of integration.]

= 2 [- cos x ]0π/3 + 3 [sin x ]0π/3

= 2 [- cos π / 3 + cos 0] + 3 [sin π / 3 - sin 0]
[Apply the limits of integration.]

= 2 [- 1 / 2 + 1] + 3 [32]

= 1 + 332 = 2+332

So, ∫0π/3(2 sin x + 3 cos x) dx = 2+332

Correct answer : (4)

Use fundamental theorem of calculus, to evaluate

\int_{0}^{1}

\frac{3}{\sqrt{4 - x^{2}}}

dx

	a.		$\frac{1}{2}$
	b.		$π$
	c.		$\frac{π}{2}$
	d.		$\frac{π}{6}$
	e.		$\frac{3π}{2}$

Solution:

Let x = 2 sin θ, dx = 2 cosθ dθ

At x = 0, θ = 0 and at x = 1, θ = π6.

∫0134-x2 dx

= ∫0π/634-4sin2θ(2 cosθ) dθ

= ∫0π/66cosθ2cosθ dθ

= ∫0π/63 dθ

= 3 [ θ ]0π/6
[Apply the limits of integration.]

= 3π6 = π2

So, ∫0134-x2 dx = π2

Correct answer : (3)

Use the fundamental theorem of calculus, to evaluate

\int_{0}^{π}

x

cos²

x

dx

	a.		$π^{2}$
	b.		$\frac{π^{2} - 1}{4}$
	c.		$\frac{π^{2} + 1}{4}$
	d.		$\frac{π^{2}}{4}$
	e.		$\frac{1}{4}$

Solution:

∫0πx cos²x dx

= ∫0πx[1+cos2x2] dx

= ∫0πx2 dx + ∫0πxcos2x2 dx
[Use Sum Rule of integration.]

= [x24 ]0π + ∫0πx cos2x2 dx

= π24 + 1 / 2 [[xsin2x2 ]0π - 1 / 2 ∫0πsin 2x dx]
[Use integration by parts.]

= π24 + [xsin2x4 ]0π - 1 / 4 [-cos2x2 ]0π

= π24 + πsin2π4 + 1 / 8 [cos 2π - cos 0]
[Apply the limits of integration.]

= π24 + 0 + 1 / 8 [1 - 1]

= π24

So, ∫0πx cos²x dx = π24

Correct answer : (4)

Use the fundamental theorem of calculus to evaluate

\int_{0}^{1}

\frac{x^{2} - 1}{x^{2} + 1}

dx

	a.		$\frac{2 + π}{2}$
	b.		1
	c.		$\frac{π}{4}$
	d.		$\frac{2 - π}{2}$
	e.		$\frac{1 - π}{4}$

Solution:

∫01x2-1x2+1 dx

= ∫01x2+1-2x2+1 dx
[Rearrange the numerator.]

= ∫01x2+1x2+1 dx - ∫012x2+1 dx
[Use Difference Rule of integration.]

= ∫01dx - 2 ∫011x2+12 dx

= [ x ]01 - 2 [tan-1(x1) ]01
[Use ∫1x2+a2 dx = 1atan-1(xa) + C.]

= 1 - 2Tan ^-1(1)
[Apply the limits of integration.]

= 1 - π2 = 2-π2

So, ∫01x2-1x2+1 dx = 2-π2

Correct answer : (4)

Use the fundamental theorem of calculus to evaluate

\int_{-3}^{0}

\frac{x}{x^{2} + 5}

dx

	a.		- $\frac{1}{2}$ ln ( $\frac{5}{14}$ )
	b.		ln ( $\frac{5}{14}$ )
	c.		$\frac{1}{2}$ ln (14)
	d.		- $\frac{1}{2}$ ln (7)
	e.		$\frac{1}{2}$ ln ( $\frac{5}{14}$ )

Solution:

∫-30 xx2+5 dx

= 1 / 2 ∫-30 2xx2+5 dx

= 1 / 2 [ln | x² + 5 | ]-30
[Use ∫f′(x)f(x) dx = ln | f(x) | + C.]

= 1 / 2 [ ln 5 - ln 14 ]
[Apply the limits of integration.]

= 1 / 2 ln (5 / 14)

So, ∫-30 xx2+5 dx = 1 / 2 ln (5 / 14)

Correct answer : (5)

Use fundamental theorem of calculus to evaluate

\int_{1}^{2}

\frac{1}{\sqrt[3]{1 - 2 x}}

dx

	a.		$\frac{3}{4}$
	b.		- $\frac{3}{4}$ ( $3^{\frac{2}{3}}$ - 1 )
	c.		$\frac{3}{4}$ ( $3^{\frac{2}{3}}$ + 1 )
	d.		( $3^{\frac{2}{3}}$ - 1 )
	e.		$\frac{1}{2}$

Solution:

Let 1 - 2x = t, - 2dx = dt, dx = - 1 / 2 dt

At x = 1, t = - 1 and at x = 2, t = - 3

∫1211-2x3 dx = - 1 / 2 ∫-1 - 31t13 dt

= - 1 / 2 ∫-1 - 3 t-13 dt

= - 1 / 2 [t2323 ]-1- 3
[Apply the limits of integration.]

= - 3 / 4 [(-3)23-(-1)23]

= - 3 / 4 ( 323 - 1)

So, ∫1211-2x3 dx = - 3 / 4 ( 323 - 1 )

Correct answer : (2)

10.

Use fundamental theorem of calculus, to evaluate

\int_{0}^{1}

x

²(2

x

- 1)⁸

dx

	a.		$\frac{1}{11}$
	b.		$\frac{20}{99}$
	c.		$\frac{2}{11}$
	d.		$\frac{1}{9}$
	e.		$\frac{2}{99}$

Solution:

Let 2x - 1 = t, 2 dx = dt, dx = dt2

At x = 0, t = - 1 and at x = 1, t = 1

∫014x²(2x - 1)⁸ dx = ∫-114(t+12)2t8dt2

= 1 / 2 ∫-11(t + 1)²t⁸ dt

= 1 / 2 ∫-11 (t² + 2t + 1)t⁸ dt

= 1 / 2 ∫-11 (t¹⁰ + 2t⁹ + t⁸) dt

= 1 / 2 ∫-11 t¹⁰ dt + ∫-11 t⁹ dt + 1 / 2 ∫-11 t⁸ dt

= 1 / 2 [t1111 ]-11 + [t1010 ]-11 + 1 / 2 [t99 ]-11

= 1 / 2 [1 / 11 + 1 / 11] + [1 / 10 - 1 / 10] + 12[19+19]
[Apply the limits of integration.]

= 1 / 11 + 1 / 9

= 20 / 99

So, ∫014x²(2x - 1)⁸ dx = 20 / 99

Correct answer : (2)